\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
Notice that \begin {align*} \DiffFrac {y_n}{x} & = \DiffFrac {(-1)^n \frac {1}{z} \NdiffFrac {n}{z}{x}}{x} \\ & = (-1)^n \left [\DiffFrac {\frac {1}{z}}{x} \cdot \NdiffFrac {n}{z}{x} + \frac {1}{z} \cdot \DiffFrac {\NdiffFrac {n}{z}{x}}{x}\right ] \\ & = (-1)^n \left [\frac {2x}{z} \cdot \NdiffFrac {n}{z}{x} + \frac {1}{z} \cdot \NdiffFrac {n + 1}{z}{x}\right ] \\ & = 2x \cdot (-1)^n \frac {1}{z} \NdiffFrac {n}{z}{x} - (-1)^{n + 1} \frac {1}{z} \NdiffFrac {n + 1}{z}{x} \\ & = 2x y_n - y_{n + 1}, \end {align*}
as desired.
We first look at the base case where \(n = 1\). What is desired is \[ y_2 = 2 x y_1 - 2 y_0. \]
We have \(y_0 = 1\), \[ y_1 = (-1)^1 \frac {1}{e^{-x^2}} \DiffFrac {e^{-x^2}}{x} = - e^{x^2} (-2x) e^{-x^2} = 2x, \] and \[ y_2 = 2x y_1 - \DiffFrac {y_1}{x} = 2x \cdot 2x - 2 = 4x^2 - 2. \]
Hence, \[ 2x y_1 - 2 y_0 = 2x \cdot 2x - 2 \cdot 1 - 4x^2 - 2 = y_2, \] so the base case is satisfied.
Now assume this is true for some \(n = k \geq 1\), i.e. \[ y_{k + 1} = 2 x y_k - 2 k y_{k - 1}. \]
We have \begin {align*} y_{k + 2} & = 2 x y_{k + 1} - \DiffFrac {y_{k + 1}}{x} \\ & = 2 x y_{k + 1} - \DiffFrac {\left (2 x y_k - 2 k y_{k - 1}\right )}{x} \\ & = 2 x y_{k + 1} - 2 y_k - 2 x \DiffFrac {y_k}{x} + 2k \DiffFrac {y_{k - 1}}{x} \\ & = 2 x y_{k + 1} - 2 y_k - 2 x (2 x y_k - y_{k + 1}) + 2k (2x y_{k - 1} - y_k) \\ & = 2 x y_{k + 1} - 2 y_k - 4x^2 y_k + 2 x y_{k + 1} + 4 k x y_{k - 1} - 2k y_k \\ & = 4x y_{k + 1} - 2 (2x^2 + k + 1) y_k + 4 k x y_{k - 1} \\ & = 4x y_{k + 1} - 2 (2x^2 + k + 1) y_k + 4 k x \cdot \frac {2 x y_k - y_{k + 1}}{2k} \\ & = 4x y_{k + 1} - 2 (2x^2 + k + 1) y_k + 2x (2 x y_k - y_{k + 1}) \\ & = 2x y_{k + 1} - 2 (k + 1) y_k, \end {align*}
which is exactly the statement for \(n = k + 1\).
Hence, by the principle of mathematical induction, we have \(y_{n + 1} = 2 x y_n - 2n y_{n - 1}\) for all \(n \geq 1\).
We have \begin {align*} \LHS & = y_{n + 1}^2 - y_n y_{n + 2} \\ & = y_{n + 1}^2 - y_n (2 x y_{n + 1} - 2 (n + 1) y_n) \\ & = y_{n + 1}^2 - 2 x y_n y_{n + 1} + 2 (n + 1) y_n^2 \end {align*}
and \begin {align*} \RHS & = 2n (y_n^2 - y_{n - 1} y_{n + 1}) + 2 y_n^2 \\ & = 2n \left (y_n^2 - \frac {2x y_n - y_{n + 1}}{2n} y_{n + 1}\right ) + 2 y_n^2 \\ & = 2n y_n^2 - \left (2x y_n - y_{n + 1}\right ) y_{n + 1} + 2 y_n^2 \\ & = 2n y_n^2 - 2x y_n y_{n + 1} + y_{n + 1}^2 + 2 y_n^2 \\ & = y_{n + 1}^2 - 2x y_n y_{n + 1} + 2 (n + 1) y_n^2. \end {align*}
This can be shown by induction on \(n\). The base case for \(n = 1\) is \[ y_1^2 - y_0 y_2 = (2x)^2 - 1 \cdot (4x^2 - 2) = 2 > 0 \] is true.
Now assume the statement is true for \(n = k \geq 1\), i.e. \[ y_k^2 - y_{k - 1} y_{k + 1} > 0. \]
We have \begin {align*} y_{k + 1}^2 - y_k y_{k + 2} & = 2n (y_k^2 - y_{k - 1} y_k + 1) + 2 y_n^2 \\ & > 2n \cdot 0 + y_n^2 \\ & = 0 + y_n^2 \\ & \geq 0, \end {align*}
which is the statement for \(n = k + 1\).
Hence, by the principle of mathematical induction, we have \(y_n^2 - y_{n - 1} y_{n + 1} > 0\) for all \(n \geq 1\).